Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), $x$ seconds after Amir threw it, is modeled by: $h(x)=-(x-2)^2+16$ How many seconds after being thrown will the ball hit the ground?
Answer: The ball hits the ground when $h(x)=0$. $\begin{aligned} h(x)&=0 \\\\ -(x-2)^2+16&=0 \\\\ -(x-2)^2&=-16 \\\\ (x-2)^2&=16 \\\\ \sqrt{(x-2)^2}&=\sqrt{16} \\\\ x-2&=\pm4 \\\\ x&=\pm4+2 \\\\ x=6&\text{ or }x=-2 \end{aligned}$ We found that $h(x)=0$ for $x=6$ or $x=-2$. Since $x=-2$ doesn't make sense in our context, the only reasonable answer is $x=6$. In conclusion, the ball will hit the ground after $6$ seconds.